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CAT 2025 Lesson : Functions - Piecewise & Non-Standard Input

6.6 Piecewise Functions

In these type of questions, function provides different outputs for different ranges of inputs. A good way to solve these is to form a table and/or assume values across different ranges.

Example 18 & 19

$f_1(x) = x$ , if $0 \le x \le 1$	$f_2(x) = f_1( - x)$ for all $x$
$= 1$ , if $x \ge 1$	$f_3(x) = - f_2(x)$ for all $x$
$= 0$ otherwise	$f_4 = f_3( - x)$ for all $x$

[CAT 2004]

How many of the following products are necessarily zero for every

x - f_4(x)f_2(x), f_2(x)f_3(x), f_2(x)f_4(x)

?

(

1

) 0 (

2

) 1 (

3

) 2 (

4

) 3

Solution

f_1(x)

is a piecewise function with

3

different outputs across

3

different ranges, we shall form a table with the

3

ranges in the columns and take

3

reference values for each range.

Note: If a positive and negative values are to be taken, avoid taking values such as

+2

and

-2

wherein each values' modulus give the same result. The of -

2

0.5

and

3

that have been used are quite distinct.

	$\bm{x \lt 0}$	$\bm{ 0 \leq x \leq 1}$	$\bm{x \gt 1}$
Reference Value	$x = -2$	$x = 0.5$	$x = 3$
$f_1(x)$	$0$	$1$	$3$
$f_2(x) = f_1(-x)$	$2$	$0$	$0$
$f_3(x) = -f_2(x)$	$-2$	$0$	$0$
$f_4(x) = f_3(-x) = -f_2(-x) = -f_1(x)$	$0$	$-1$	$-3$

f_2(x)f_3(x)

x = -2

, provides a value of

2 \times - 2 = -4

f_1(x)f_2(x)

and

f_2(x)f_4(x)

across the

3

ranges is always

0

. Therefore,

\bm{2}

of the products are always

0

.

Answer: (

3

)

2

19

) Which of the following is necessarily true?

(

1

)

f_4(x) = f_1(x)

for all

x

(

2

)

f_1(x) = -f_3(-x)

for all

x

(

3

)

f_2(-x) = f_4(x)

for all

x

(

4

)

f_1(x) + f_3(x) = 0

for all

x

Option (1) is not true at

x = 0.5

and

x = 3

.

Option (2):

-f_3(-x) = -(-f_2(-x)) = f_2(-x) = f_1(-(-x)) = f_1(x)

. This is always true.

Option (3): We note from the table that

f_4(x) = f_3(-x) = -f_2(-x)

. Therefore, this is also incorrect.

Option (4) is not true for any of the

x

-values.

Answer: (

2

)

f_1(x) = -f_3(-x)

for all

x

6.7 Non-standard input for the function

In these questions equate the required input to the function's input, solve for

x

and the substitute the

x

-value in the output.

Example 20

f (\dfrac{2x + 3}{x - 1}) = \dfrac{x +4 }{x - 2 }

, then

f(5)

= ?

Solution

In these question, we first equate the input term to

5

, solve for

x

and then substitute

x

in the output.

If

\left(\dfrac{2x + 3}{ x - 1}\right) = 5

⇒

2x + 3 = 5x - 5

⇒

x = \dfrac{8}{3}

The question states

\left(\dfrac{2x +3}{x - 1}\right) = \dfrac{x + 4}{x - 2}

When

x = \dfrac{8}{3}

f \left(\dfrac{2x + 3}{x - 1}\right) = f(5) = \dfrac{\dfrac{8}{3} + 4}{\dfrac{8}{3} - 2} = \dfrac{\dfrac{20}{3}}{\dfrac{2}{3}}

⇒

f(5) = 10

Answer:

10

Example 21

3 f(x + 2) + 4f( \dfrac{1}{x + 2}) = 4x, x \neq - 2

, the

f(4)

is
[XAT 2008]

(

1

)

7

(

2

)

\dfrac{52}{7}

(

3

)

8

(

4

)

\dfrac{56}{7}

(

5

) None of the above

Solution

The

2

functions have non-standard inputs with

1

being the reciprocal of the other. The equation looks like a linear equation with

2

variables. We need to form

2

equations to solve it.

If the first input equals

4

, i.e.

x + 2 = 4

⇒

x = 2

Substituting

x = 2

, we get

3 f(4) + 4f(\dfrac{1}{4}) = 8

-----(a)

If the second input equals

4

, i.e.

\dfrac{1}{x + 2} = 4

⇒

x = - \dfrac{7}{4}

Substituting

x

=

- \dfrac{7}{4}

, we get

3 f\left(\dfrac{1}{4}\right) + 4f(4) = -7

-----(b)

4 \times (b) - 3 \times (a)

⇒

4f(4) = - 52

⇒

f(4)

=

-

\dfrac{52}{7}

Answer: (

5

) None of the above

6.8 Substitute and get the answer

In these questions, substitute small integral values for

x

, starting from

1

and find the option that is true.

Example 22

f(x) = \dfrac{3x + 2}{2x - 3}

, then which of the following is true?

(

1

)

f(x) = f(-x)

(

2

)

f(x) = \dfrac{-1}{f(-x)}

(

3

)

x = f(f(x))

(

4

)

f(x) = -f(-x)

Solution

In these questions, substitute values for

x

and proceed by eliminating the answer options.

f(1) = \dfrac{3 + 2}{2 - 3} = -5

and

f(-1) = \dfrac{-3 + 2}{-2 - 3} =\dfrac{1}{5}

f(1) \neq f(-1)

, option (

1

) is rejected
As

f(1) \neq -f(-1)

, option (

4

) is rejected

As

f(1) = \dfrac{-1}{f(-1)}

, we keep option (

2

) on hold and check if option (

3

) can be rejected.

f(f(1)) = f(-5) = \dfrac{-15 + 2}{-10 - 3} = 1

,

∴ Option (

3

) cannot be rejected. We need to substitute a different value, say

x = 2

and check for options

2

and

3

alone.

f(2) = 8

and

f(-2) = \dfrac{4}{7}

Clearly,

f(2) \neq \dfrac{-1}{f(-2)}

and option (

2

) is rejected. And, the correct answer is option ( $\bm{3}$ ).

Note that

f(f(2)) = f(8) = \dfrac{24 + 2}{16 - 3} = 2

. So, option (

3

) is true.

Answer: (

3

)

x = f(f(x))

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